## - Argand Diagram

A complex number $$z=a+bi$$ can be represented on a diagram called an Argand diagram as

(i) a point with coordinates $$(a, b)$$
(ii) a vector

REPRESENTING SUMS AND DIFFERENCES ON ARGAND DIAGRAMS
LESSON 1
Find $$z_1+z_2$$ and $$z_1-z_2$$ for $$z_1=6+2i$$ and $$z_2=4+9i$$. Hence, represent $$z_1+z_2$$ and $$z_1-z_2$$ on Argand diagrams.
SOLUTION
(i) $$z_1+z_2=(6+2i)+(4+9i)=10+11i$$
Letting $$z_3=z_1+z_2$$ we see that $$z_3$$ is the diagonal of the parallelogram formed with $$z_1$$ and $$z_2$$ as sides its sides.

(ii) $$z_1-z_2=(6+2i)-(4+9i)=2-7i$$
Letting $$z_3=z_1-z_2$$ we see that in the case of the subtraction of complex numbers that $$z_3$$ and $$z_2$$ represent the sides of the parallelogram with $$z_1$$ being one of the diagonals.

THE MODULUS OF A COMPLEX NUMBER
The modulus of a complex number, $$z=a+bi$$, is a measure of the magnitude of $$z$$, and is written as $$|z|$$.
Thus modulus $$z=|z|=\sqrt{a^2+b^2}$$.

LESSON 1
Determine the modulus of
(i) $$z_1=1+i$$
(ii) $$z_2=-3+4i$$
(iii) $$z_3=-1-\sqrt{3}i$$
(iv) $$z_4=-5$$
SOLUTION
(i) $$z_1=1+i$$
$$|z_1 |=\sqrt{1^2+1^2}=\sqrt{2}$$

(ii) $$z_2=-3+4i$$
$$|z_2 |=|-3+4i|=\sqrt{(-3)^2+4^2}=5$$

(iii) $$z_3=-1-\sqrt{3} i$$
$$|z_3 |=|-1-\sqrt{3} i|=\sqrt{(-1)^2+(-\sqrt{3})^2}=2$$

(iv) $$z_4=-5$$
$$|z_4 |=5$$

LESSON 2
If $$z_1=-3+4i$$ and $$z_2=2-i$$, what is $$|z_1-z_2|$$?
SOLUTION
We are trying to find the distance between $$z_1$$ and $$z_2$$. In other words, what is the distance between the points, say, $$P(-3, 4)$$ and $$Q(2, -1)$$ on the Argand Diagram?

Since we know the coordinates of the two points we can use the formula for the length of a line segment.
$$|PQ|=\sqrt{(x_2-x_1 )^2+(y_2-y_1 )^2}$$
$$⟹|z_1-z_2 |=\sqrt{(-3-2)^2+(4-(-1))^2}$$
$$=\sqrt{50}$$

THE ARGUMENT OF COMPLEX NUMBER
The angle θ is called the argument of $$z (\arg ⁡z)$$ where $$\theta$$ is the angle the vector representing the complex number on the Argand diagram makes with the positive real axis. Thus $$\displaystyle \tan⁡θ={b\over a}$$. To avoid complications, we use $$–π<θ≤π$$ and this is known as the principal argument of $$z$$.

LESSON 1
Determine the argument of
(a) $$z_1=1+i$$
(b) $$z_2=-3+4i$$
(c) $$z_3=-1-\sqrt{3} i$$
(d) $$z_4=-5$$
SOLUTION
It is useful to draw a sketch of the complex number so that you know which quadrant the complex number is in and then you can determine the appropriate formula for the argument.
(a) $$z=1+i$$
$$z$$ is in QUADRANT 1 since both the real and imaginary parts are positive. Therefore $$\displaystyle \arg ⁡z=\tan^{-1}\left({b\over a}\right)$$
$$\arg⁡ (z_1)=θ$$
$$\displaystyle =\tan^{-1}\left({1\over1}\right)$$
$$\displaystyle ={π\over4}$$

(b) $$z=-3+4i$$
$$z$$ is in Quadrant II since the real part is negative and the imaginary part is positive. Therefore, $$\displaystyle \arg ⁡z=π-\tan^{-1}⁡\left({b\over a}\right)$$
$$\arg⁡ (z_2)=θ$$
$$\displaystyle =π-\tan^{-1}\left({4\over3}\right)$$
$$\displaystyle =2.21$$

(c) $$z=-1-\sqrt{3} i$$
Both the imaginary and real parts for $$z$$ are negative. This implies that $$z$$ is in quadrant IV, $$\displaystyle \arg ⁡z=-π+\tan^{-1}\left({b\over a}\right)$$.
$$\arg⁡ (z_3)=θ$$
$$\displaystyle =-π+\tan^{-1}⁡\left({\sqrt{3}\over 1}\right)$$
$$\displaystyle =-{2π\over3}$$
(d) $$z_4=-5$$
This is simply a straight line on the $$x$$ – axis.
$$\arg⁡ (z_4)=π$$

MODULUS – ARGUMENT FORM

From the diagram
$$\displaystyle \sin ⁡θ={b\over r}$$
$$⟹r \sin ⁡θ=b$$

$$\displaystyle \cos⁡ θ={a\over r}$$
$$r \cos⁡θ=a$$

Given that
$$z=a+bi$$
replacing $$a$$ and $$b$$ we get the modulus argument form of a complex number
$$⟹z=r \cos⁡θ+ri \sin⁡θ$$
$$⟹z=r(\cos⁡θ+i \sin⁡θ)$$

LESSON 1
Write the following in modulus – argument form
(a) $$z_1=1+i$$
(b) $$z_2=-3+4i$$
(c) $$z_3=-1-\sqrt{3} i$$
(d) $$z_4=-5$$
SOLUTION
The moduli and arguments for these complex numbers were determined previously. Therefore, the required modulus – argument form is shown directly.
(a) $$z_1=1+i$$
$$\displaystyle z_1=\sqrt{2} \left(\cos⁡ \left({π\over4}\right)+i \sin \left({π\over4}\right)\right)$$

(b) $$z_2=-3+4i$$
$$z_2=5(\cos⁡(2.21)+i \sin⁡(2.21))$$

(c) $$z_3=-1-\sqrt{3} i$$
$$\displaystyle z_3=2\left(\cos\left(-{2π\over3}\right)+i\sin\left(-{2π\over3}\right)\right)$$

(d) $$z_4=-5$$
$$z_4=5(\cos ⁡π+i \sin ⁡π)$$

LESSON 2
Given that $$z_1=1+\sqrt{3} i$$ and $$z_2=1-i$$, determine (i) $$z_1 z_2$$ and (ii) $$\displaystyle {z_1\over z_2}$$ in modulus – argument form.
SOLUTION
• Determine the modulus and principal argument of $$z_1$$
$$|z_1 |=\sqrt{1^2+(\sqrt{3})^2}=2$$
$$\displaystyle \arg⁡ (z_1)=\tan^{-1}\left({\sqrt{3}\over1}\right)={π\over3}$$
• Determine the modulus and principal argument of $$z_2$$
$$|z_2 |=\sqrt{1^2+1^2}=\sqrt{2}$$
$$\displaystyle \arg⁡ (z_2)=-\tan^{-1}⁡\left({1\over1}\right)={π\over4}$$
• Use the appropriate formulae to determine the required solutions.
$$z_1 z_2=r_1 r_2 [\cos⁡(A+B)+i \sin⁡(A+B)]$$
$$\displaystyle z_1 z_2=2\sqrt{2} \left[\cos\left({π\over3}+{π\over4}\right)+i \sin\left({π\over3}+{π\over4}\right) \right]$$
$$\displaystyle =2\sqrt{2} \left[\cos⁡\left({7π\over12}\right)+i \sin⁡\left({7π\over12}\right)\right]$$

$$\displaystyle {z_1\over z_2}={r_1\over r_2}\left[\cos⁡(A-B)+i \sin⁡(A-B)\right]$$
$$\displaystyle {z_1\over z_2} ={2\over \sqrt{2}} \left[\cos\left({π\over3}-{π\over4}\right)+i \sin\left({π\over3}-{π\over4}\right)\right]$$
$$\displaystyle =\sqrt{2} \left[\cos⁡ \left({π\over12}\right)+i \sin⁡\left({π\over12}\right)\right]$$